What kind of object is it in $\mathbb{R}^4$

Question

Given an object in $\mathbb{R}^4$ : $\{(x_1,x_2,x_3, x_4)\in\mathbb{R}^4:x_1^2+x_2^2 =\frac{1}{2} , x_3^2+x_4^2 =\frac{1}{2}\}$ What kind of object could it be? How could I determine its parametric equation $(x_1,x_2,x_3, x_4)=(f_1(t),f_2(t),f_3(t),f_4(t))$?

Answer 1

This object is called "flat torus". And it is inside a sphere of radius 1.

A parametrization for it is:

$$X(\theta,\phi)=(\frac{1}{\sqrt{2}}\cos\theta,\frac{1}{\sqrt{2}}\sin\theta,\frac{1}{\sqrt{2}}\cos\phi,\frac{1}{\sqrt{2}}\sin\phi).$$

Answer 2

generic hint: Consider $\mathbb R^4$ as $\mathbb R^2 \times \mathbb R^2$. What does the set look like here?

If you call the subset $A$, consider the map $g:I^2 \to A \subset \mathbb R^4$ given by $t,s \mapsto (\frac{1}{2}e^{it},\frac{1}{2}e^{is})$. I've written it this way to emphasize that the first two co-ordinates sit on a circle, and the latter two sit on another circle, both of which are embedded in $\mathbb C \cong \mathbb R^2$.

geometric: the latter map imposes some kind of equivalence relation on $I^2$. What is it? Is this a familiar object after gluing?

analytic:

$A$ is the zero set of a function $F:\mathbb R^4 \to \mathbb R^2$ given by $(x,y,z,w) \mapsto (x^2+y^2-1/2,z^2+w^z-1/2)$.

If you parametrize by the first map (replacing $e^{it}$ with $\sin(t),\cos(t)$ and likewise for $s$) you will have thaat $F \circ g:\mathbb R^2 \to \mathbb R^2$. This should be the zero map, which implies that the parametrization is a subset of $A$. Surjectivity is easy.